Question

find v.
right triangle with angles 30°, 60°, 90°, side opposite 30° is 5 mi, side v is adjacent to 30°
write your answer in simplest radical form.
blank miles
submit button

Explanation:

Step1: Identify triangle type

This is a 30-60-90 right triangle. In such a triangle, the sides are in the ratio \(1 : \sqrt{3} : 2\) (opposite 30°, 60°, 90° respectively).

Step2: Relate given side to ratio

The side of length 5 mi is opposite the 30° angle? Wait, no. Wait, the right angle, 30°, 60°: the side opposite 30° is the shortest, opposite 60° is \( \sqrt{3} \) times that, and hypotenuse is twice the shortest. Wait, in the triangle, the side labeled 5 mi: let's check angles. The angle of 30°: the side adjacent to 30°? Wait, no, let's use trigonometry. Let's see, angle 30°, right angle, so the side opposite 30° is... Wait, the side with length 5 mi: let's see the angles. The angle at the bottom is 60°, so the side opposite 60° is \( v \)? Wait, no. Wait, the right angle is at the top right. So the sides: let's label the triangle. Let’s call the right angle \( C \), 30° angle \( A \), 60° angle \( B \). Then side opposite \( A \) (30°) is \( BC \), opposite \( B \) (60°) is \( AC \), and hypotenuse \( AB \). Wait, the given side is 5 mi: let's see, the side labeled 5 mi is adjacent to 60°? Wait, maybe better to use sine or cosine. Let's use sine of 30°: \( \sin(30°) = \frac{\text{opposite}}{\text{hypotenuse}} \). Wait, no, the side of 5 mi: wait, maybe the side of 5 mi is opposite the 30° angle? Wait, no, 30° angle's opposite side would be shorter. Wait, no, in a 30-60-90 triangle, the sides are \( x \), \( x\sqrt{3} \), \( 2x \), where \( x \) is opposite 30°, \( x\sqrt{3} \) opposite 60°, \( 2x \) hypotenuse. Wait, looking at the triangle, the side with length 5 mi: let's see, the angle at the bottom is 60°, so the side opposite 60° is \( v \), and the side opposite 30° is 5 mi? Wait, no, that can't be. Wait, no, let's check the angles. The angle at the top left is 30°, right angle at top right, so the side between 30° and right angle is \( v \) (adjacent to 30°), the side between right angle and 60° is 5 mi (opposite to 30°). Wait, yes! So the side opposite 30° is 5 mi? Wait, no, 30° angle's opposite side is the shortest. Wait, no, if the side opposite 30° is 5 mi, then the hypotenuse is \( 2 \times 5 = 10 \), and the other leg (opposite 60°) is \( 5\sqrt{3} \). But wait, the side we need to find is \( v \), which is adjacent to 30° and opposite to 60°. Wait, maybe I mixed up. Let's start over. In a right triangle, with angles 30°, 60°, 90°: - The side opposite 30° is the shortest side, let's call it \( x \). - The side opposite 60° is \( x\sqrt{3} \). - The hypotenuse is \( 2x \). Now, looking at the triangle: the side labeled 5 mi: is it the side opposite 30°? If so, then \( x = 5 \), so the side opposite 60° (which is \( v \)) would be \( x\sqrt{3} = 5\sqrt{3} \)? Wait, no, that doesn't match. Wait, maybe the side of 5 mi is the side opposite 60°? No, that would be longer. Wait, maybe the side \( v \) is the hypotenuse? No, the hypotenuse is the longest side. Wait, let's use trigonometry. Let's take angle 30°: \( \cos(30°) = \frac{\text{adjacent}}{\text{hypotenuse}} \). Wait, the adjacent side to 30° is \( v \), and the opposite side is 5 mi? Wait, no, the right angle is at the top right, so the sides: horizontal side is \( v \) (adjacent to 30°), vertical side is 5 mi (opposite to 30°). Then \( \tan(30°) = \frac{\text{opposite}}{\text{adjacent}} = \frac{5}{v} \). So \( \tan(30°) = \frac{1}{\sqrt{3}} = \frac{5}{v} \), so \( v = 5\sqrt{3} \)? Wait, no, that would be if 5 is opposite 30°. Wait, no, \( \tan(30°) = \frac{\text{opposite}}{\text{adjacent}} \), so if 30° angle is at the top left, then the opposite side is the vertical side (5 mi), and adjacent is horizontal side \( v \). So \( \tan(30°) = \frac{5}{v} \), so \( v = \frac{5}{\tan(30°)} \). Since \( \tan(30°) = \frac{1}{\sqrt{3}} \), so \( v = 5\sqrt{3} \). Wait, but let's check with sine. \( \sin(60°) = \frac{\text{opposite}}{\text{hypotenuse}} \). Wait, no, maybe the side \( v \) is opposite 60°, and the hypotenuse is something else. Wait, alternatively, use the 30-60-90 ratios. If the side opposite 30° is 5 mi, then the side opposite 60° (which is \( v \)) is \( 5\sqrt{3} \), and hypotenuse is 10. But wait, is the 5 mi side opposite 30°? Let's see the angles: the 30° angle is at the top left, so the side opposite to it is the vertical side (5 mi), yes. So then the side opposite 60° (which is \( v \)) is \( 5\sqrt{3} \). Wait, but let's confirm with cosine. \( \cos(30°) = \frac{\text{adjacent}}{\text{hypotenuse}} \), adjacent is \( v \), hypotenuse is 10 (since hypotenuse is twice the opposite of 30°). So \( \cos(30°) = \frac{v}{10} \), so \( v = 10 \times \cos(30°) = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3} \). Yes, that matches. So \( v = 5\sqrt{3} \).

Answer:

\( 5\sqrt{3} \)